Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at 50∘C has a total translational kinetic energy of 4000 J.

A) (Multiple Choice) Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50∘C is:

          a) (16)(2000m/s)=32000m/s

          b) (4)(2000m/s)=8000m/s

          c) 2000m/s

          d) (14)(2000m/s)=500m/s

          e) (116)(2000m/s)=125m/s

           f) none of the above

B) The total translational kinetic energy of 1.0 mole of diatomic oxygen at 50∘C is:

a) (16)(4000J)=64000J
b)	(4)(4000J)=16000J
c)	4000J
d)	(14)(4000J)=1000J
e)	(116)(4000J)=150J
f)	none of the above

C) The temperature of the diatomic hydrogen gas sample is increased to 100∘C. The root-mean-square speed vrms for diatomic hydrogen at 100∘C is:

a) (2)(2000m/s)=4000m/s
b)	(2√)(2000m/s)=2800m/s
c)	2000m/s
d)	(12√)(2000m/s)=1400m/s
e)	(12)(2000m/s)=1000m/s
f)	none of the above

Answer 1

Concepts and reason

The concept used to solve this problem is Root mean square speed and translational kinetic energy of a molecule.

First find the value of root mean square speed and using this find translational kinetic energy. And form the definition of the root mean square velocity the rms speed depends on the temperature and mass of the system.

Fundamentals

Root mean square speed of a particle is expressed as follows:

${v_{rms}} = \sqrt {\frac{{3RT}}{m}}$ …… (1)

Here, R is universal gas constant having value ${\rm{8}}{\rm{.31}} {\rm{J/mol/K}}$, $T$ is provided temperature in Kelvin and$m$is the molecular mass of gas,

Translational kinetic energy is the energy of a gas molecule due to its linear motion and is expressed as follows:

$T = \frac{1}{2}mv_{rms}^2$ …… (2)

Here, $m$ is the molecular mass of gas and ${v_{{\rm{rms}}}}$ is the root mean square speed of the molecule.

(A)

Molecular mass of diatomic hydrogen is ${\rm{2 g/mol}}$. Diatomic oxygen has molecular mass 16 times that of diatomic hydrogen, so molecular mass of diatomic oxygen is $\begin{array}{c}\\{\rm{Molecular mass of diatomic oxygen}} = {\rm{32}} {\rm{g/mol }}\\\\ = {\rm{3}}\left( {\frac{{\rm{g}}}{{{\rm{mol}}}}} \right)\left( {\frac{{{\rm{1 kg}}}}{{{\rm{1000 g}}}}} \right)\\\\ = 0.032 {\rm{kg/mol}}\\\end{array}$.

Convert temperature given in Celsius to kelvin using formula.

${T_K} = {{\rm T}_{^C}} + 273.15$

Here, ${T_K}$ is temperature in Kelvin and ${{\rm T}_{^C}}$is temperature in Celsius.

Substituting ${50^ \circ } {\rm{C}}$for ${{\rm{T}}_{\rm{C}}}$in the above formula,

$\begin{array}{c}\\{T_K} = \left( {{\rm{5}}{{\rm{0}}^{\rm{o}}}{\rm{C}} + {\rm{ 273}}{\rm{.15}}} \right) {\rm{K}}\\\\ = {\rm{373}}{\rm{.15 K }}\\\end{array}$.

Substituting ${\rm{8}}{\rm{.31}} {\rm{J/mol/K}}$for$R$, $373 {\rm K}$for $T$and ${\rm{0}}{\rm{.032 kg}}$for $m$in equation (1) .$\begin{array}{c}\\{v_{rms}} = \sqrt {\frac{{3RT}}{m}} \\\\ = \sqrt {\frac{{3\left( {{\rm{8}}{\rm{.31}} {\rm{J/mol/K}}} \right)\left( {373 {\rm K}} \right)}}{{{\rm{0}}{\rm{.032 kg}}}}} \\\\ = \sqrt {251637.1875 {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\\\ \approx 500{\rm{ m/s}}\\\end{array}$

(B)

Calculate the molecular weight of diatomic oxygen.

$\begin{array}{c}\\{\rm{Molecular weight of diatomic oxygen}} = {\rm{32}} {\rm{g/mol }}\\\\ = {\rm{32}}\left( {\frac{{\rm{g}}}{{{\rm{mol}}}}} \right)\left( {\frac{{{\rm{1 kg}}}}{{{\rm{1000 g}}}}} \right)\\\\ = 0.032 {\rm{kg/mol}}\\\end{array}$

Substituting $0.032{\rm{ kg/mol}}$for $m$and $500 {\rm{m/s}}$for ${v_{rms}}$,

$\begin{array}{c}\\{T_{K.E}} = \frac{1}{2}\left( {0.032{\rm{ kg/mol}}} \right)\left( {500 {\rm{m/s}}} \right)\left( {500 {\rm{m/s}}} \right)\\\\ = 4000 {\rm{J}}\\\end{array}$

Here, ${T_{K.E}}$is the translational kinetic energy.

(C)

Calculate the Molecular weight of diatomic hydrogen.

is $2{\rm{ g/mol}}$.

$\begin{array}{c}\\{\rm{Molecular weight of diatomic hydrogen}} = {\rm{2}} {\rm{g/mol }}\\\\ = {\rm{2}}\left( {\frac{{\rm{g}}}{{{\rm{mol}}}}} \right)\left( {\frac{{{\rm{1 kg}}}}{{{\rm{1000 g}}}}} \right)\\\\ = 0.002 {\rm{kg/mol}}\\\end{array}$

Substituting ${100^ \circ } {\rm{C}}$for ${{\rm{T}}_{\rm{C}}}$in the formula, ${T_K} = \left( {{T_C} + 273.15} \right) {\rm K}$

$\begin{array}{c}\\{T_K} = \left( {{\rm{5}}{{\rm{0}}^{\rm{o}}}{\rm{C}} + {\rm{273}}{\rm{.15}}} \right) {\rm{K}}\\\\ = {\rm{373}}{\rm{.15 K }}\\\end{array}$.

Substituting ${\rm{8}}{\rm{.31}} {\rm{J/mol/K}}$for$R$,${\rm{373 {\rm K}}}$ for $T$and $0.002 {\rm{kg/mol}}$for $m$.

$\begin{array}{c}\\{v_{rms}} = \sqrt {\frac{{{\rm{3RT}}}}{m}} \\\\ = \sqrt {\frac{{3\left( {{\rm{8}}{\rm{.31}} {\rm{J/mol/K}}} \right)\left( {{\rm{373 {\rm K}}}} \right)}}{{0.002 {\rm{kg/mol}}}}} \\\\ = 2156 {\rm{m}}/{\mathop{\rm s}\nolimits} \\\end{array}$

Ans: Part A

The Root mean square speed for diatomic oxygen at ${\bf{ 5}}{{\bf{0}}^{\bf{o}}}{\bf{C}}$ is ${\bf{500 m/s}}$.

Part B

Translational kinetic energy of 1 mol of diatomic oxygen at ${\bf{5}}{{\bf{0}}^{\bf{o}}}{\bf{ C}}$is ${\bf{4000 J}}.$

Part C

None of the answer choices are correct.